The ten’s digit of a three-digit number is 3. If the digits of the number are interchanged and the number thus formed is 396 more than the previous one, and the sum of the unit digit and hundred digit is 14, then what is the number?

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**Solution:**

1. Let the three-digit number be $xyz$, where $x$ is the hundred's digit, $y$ is the ten's digit, and $z$ is the unit digit.

2. Given that the ten's digit $y = 3$.

3. The number is $100x + 10y + z = 100x + 30 + z$.

4. When digits are interchanged, the new number formed is $100z + 10y + x = 100z + 30 + x$.

5. According to the problem, the new number is 396 more than the original number:
   $$ (100z + 30 + x) - (100x + 30 + z) = 396 $$

6. Simplify the equation:
   $$ 100z + x - 100x - z = 396 $$
   $$ 99z - 99x = 396 $$
   $$ 99(z - x) = 396 $$
   $$ z - x = \frac{396}{99} = 4 $$

7. Also given that the sum of the unit digit and hundred digit is 14:
   $$ z + x = 14 $$

8. Solve the system of equations:
   - $z - x = 4$
   - $z + x = 14$

Adding both equations:
   $$ 2z = 18 \implies z = 9 $$

Substitute $z=9$ into $z + x = 14$:
   $$ 9 + x = 14 \implies x = 5 $$

9. Therefore, the number is:
   $$ 100x + 10y + z = 100 	imes 5 + 10 	imes 3 + 9 = 500 + 30 + 9 = 539 $$

**Answer:** $539$

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